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`        Derive a relation b/w Cp and Cv for a real gas obeying vander wall's relation.`
one year ago

```							= R / (1 - [2a×(V-b)²]/[RTV³])= [(RT)/(V-b)] × [R/(V-b)] / [(RT)/(V-b)² -2a/V³]Cp - Cv = [p+a/V²] × [R/(V-b)] / [(RT)/(V-b)² -2a/V³](?V/?T)p = [R/(V-b)] / [(RT)/(V-b)² -2a/V³]=>-R/(V-b) +[-2a/V³ + (RT)/(V-b)²] (?V/?T)p = 0(p + a/V²) - (RT)/(V-b) = 0The partial derivative of the volume you get by implicit differentiation(?E/?V)t = a/V²As shown before (see link):(p + a/V²) = (RT)/(V-b)For a Van der Waals gasCp - Cv = [p + 0] × R/p = R(?V/?T)p = R/p(?E/?V)t = T×R/V - p = 0pV = RT  V = (RT)/p p = (RT)/V For an ideal gas(?E/?V)t = T×(?p/?T)v - pAs shown in my answer to your last question (see link):For further calculations the is a useful relation to evaluate the partial derivative of the energy from an equation of state.q.e.d.Cp - Cv = (?E/?V)t (?V/?T)p + p×(?V/?T)p = [(?E/?V)t + p]×(?V/?T)pJoin both expression together:(?E/?T)p = (?E/?V)t (?V/?T)p + CvThe partial derivative of E with respect to temperature at constant pressure is:dE = (?E/?V)t dV + (?E/?T)v dT = (?E/?V)t dV + Cv dTBecauseCp = (?E/?T)p + (?(pV)/?T)p = (?E/?T)p + p×(?V/?T)pTherefore:H = E + pVSubstitute the enthalpy by its definitionCp = (?H/?T)p ((..)p indicates derivation at constant pressure)Start with the definition of heat capacity at constant pressure
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one year ago
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